Find the angle between two vectors such that the magnitude of the sum and difference of two vectors are same
Answers
Answered by
0
If the magnitude of the sum of two vectors is equal to the difference in their magnitudes, then what is the angle between the vectors? ... This means the angle has to be 180. Mathematically, Let the vectors be a and b with magnitudes a and b respectively and the angle between them be x.
logically, how can magnitude of vector sum be equal to difference of their magnitudes. Obviously if the vectors are in opposite directions. This means the angle has to be 180.
Mathematically,
Let the vectors be a and b with magnitudes a and b respectively and the angle between them be x.
Magnitude of the sum of a and b is
√(a^2+b^2+2abcosx
Difference in their magnitudes is
a-b
Hence,
√(a^2+ b^2+2ab cosx) = a-b
Squaring both sides,
a^2+b^2+2ab cos x = a^2+ b^2–2ab
2ab cosx+2ab =0
2ab(cosx +1) =0
Since 2ab can't be zero,
Cos x+1=0
Cosx=-1
X=180
simple mate !!!
logically, how can magnitude of vector sum be equal to difference of their magnitudes. Obviously if the vectors are in opposite directions. This means the angle has to be 180.
Mathematically,
Let the vectors be a and b with magnitudes a and b respectively and the angle between them be x.
Magnitude of the sum of a and b is
√(a^2+b^2+2abcosx
Difference in their magnitudes is
a-b
Hence,
√(a^2+ b^2+2ab cosx) = a-b
Squaring both sides,
a^2+b^2+2ab cos x = a^2+ b^2–2ab
2ab cosx+2ab =0
2ab(cosx +1) =0
Since 2ab can't be zero,
Cos x+1=0
Cosx=-1
X=180
simple mate !!!
Answered by
0
hello,
let the vectors be A and B with an angle ∆ between them ,
according to question=>
|A+B|=|A-B| ------(1)
=> √(A^2 + B^2 +2ABcos∆)= √(A^2+B^2-2ABcos∆)
=> cos∆ =0
=> ∆ = 90° or 270° -------(as 0°< ∆<360°)
let the vectors be A and B with an angle ∆ between them ,
according to question=>
|A+B|=|A-B| ------(1)
=> √(A^2 + B^2 +2ABcos∆)= √(A^2+B^2-2ABcos∆)
=> cos∆ =0
=> ∆ = 90° or 270° -------(as 0°< ∆<360°)
Similar questions